Array-2

Find Disappeared Number.py

@@ -0,0 +1,18 @@
+# Time Complexity --> O(n)
+# Space Complxity --> O(1)
+class Solution:
+    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        re = []
+
+        for i in range(n):
+            index = abs(nums[i])-1
+            if nums[index]>0:
+                nums[index] = -1*nums[index]
+
+        for i in range(n):
+            if nums[i]>0:
+                re.append(i+1)
+            nums[i] = abs(nums[i])
+        return re 
+

GameOfLife.py

@@ -0,0 +1,44 @@
+# Time Complexity --> O(m*n)
+# Space Complexity --> O(1)
+class Solution:
+    def gameOfLife(self, board: List[List[int]]) -> None:
+        """
+        Do not return anything, modify board in-place instead.
+        """
+        # 1 --> 0 then 2
+        # 0 --> 1 then 3
+        m = len(board)
+        n = len(board[0])
+
+        def helper(r, c):
+            dim = [[0,1],[1,0],[-1,0],[0,-1],[-1,-1],[-1,1],[1,-1],[1,1]]
+
+            re = 0
+            for i in dim:
+                row = r+i[0]
+                col = c+i[1]
+                if 0<= row <m and 0<= col<n:
+                    if board[row][col]==2 or board[row][col]==1:
+                        re = re + 1
+            return re 
+
+        for i in range(m):
+            for j in range(n):
+                if board[i][j]==0 and helper(i,j)==3:
+                    board[i][j]=3
+                elif board[i][j]==1 and helper(i,j)<2:
+                    board[i][j]=2
+                elif board[i][j]==1 and helper(i,j)>3:
+                    board[i][j]=2
+                else:
+                    pass
+
+        for i in range(m):
+            for j in range(n):
+                if board[i][j]==2:
+                    board[i][j]=0
+                elif board[i][j]==3:
+                    board[i][j]=1 
+
+
+

Problem2.py

@@ -0,0 +1,26 @@
+# Time Comkplexity --> O(n)
+# Space Complexity --> O(1)
+# Total Comparisons --> 1.5*n. We are comparing a pair of numbers at a time and then comparing the minimum of them with existing min value and maximum of them with existing max value
+def findMinAndMax(self, nums):
+      n = len(nums)
+      i = 0
+
+      ## Different initializations depending on whether the length of array is even or odd
+      if n % 2 == 0:
+          min_val = min(nums[0], nums[1])
+          max_val = max(nums[0], nums[1])
+          i = 2
+      else:
+          min_val = max_val = nums[0]
+          i = 1
+
+      while i < n - 1:
+          if nums[i] < nums[i + 1]:
+              min_val = min(min_val, nums[i])
+              max_val = max(max_val, nums[i + 1])
+          else:
+              min_val = min(min_val, nums[i + 1])
+              max_val = max(max_val, nums[i])
+          i += 2
+
+      return [min_val, max_val]